A test charge 'q' is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shows in the figure. (i) Calculate the potential difference between A and C. (ii) At which point (of the two) is the electric potential more and why ?

Explanation

(i) As electric field is a conservative field, work doene in moving the test charge q is exactly same along the path ABC or along straight path AC.
Distance `AC = vec r = - 4 hati`, and force applied against the electric field `vecF_("ext")= - vecE_q = -E_qhati`
`:.` Work done `W_(AC) = vec F_("ext").vecr = [-Eq hat i].[-4hati] = + 4 Eq`
`:.` Potential difference between C and A
`V_C - V_A = (W_(AC))/q = (4Eq)/q = + 4E`
(ii) Electric potential at point C is greater i.e., `V_C gt V_A`.